Applying the sum-of-digits operation to $4444^{4444}$ three times

Asked Jul 24 '15 at 18:14

Active Jul 30 '15 at 12:52

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$A$ is the sum of the digits of $4444^{4444}$. $B$ is the sum of the digits of $A$. Then what will be the sum of digits of $B$?

I found this in the Suggestion paper of Mathematical Tripos Exam of Cambridge

edited Jul 30 '15 at 12:52

asked Jul 24 '15 at 18:14

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This is from an IMO – Asinomás Jul 24 '15 at 18:14
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Found it here! http://www.artofproblemsolving.com/wiki/index.php/1975_IMO_Problems/Problem_4 – Asinomás Jul 24 '15 at 18:15
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http://math.stackexchange.com/questions/169797/write-down-the-sum-of-sum-of-sum-of-digits-of-44444444?rq=1 – Brian Cheung Jul 24 '15 at 18:15
answer is $7$... – Asinomás Jul 24 '15 at 18:29
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Those are not bogus comment; they will give you the answer, if you just bother to follow the links and read... – Hans Lundmark Jul 24 '15 at 18:37
I know what you want, but I don't think that "sum of digits" is the correct expression. Maybe "reduced sum of digits" or "iterated sum of digits". For example, the sum of the digits of $4444$ is $16$. What you want is $7$. – Steven Alexis Gregory Jul 25 '15 at 03:22

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